#用filter()和lambda表达式快速求出100以内所有3的倍数的数
print(list(filter(lambda x: False if x % 3 else True ,range(1,100))))

print(list(filter(lambda n : not(n%3), range(1, 100))))

print(list(zip([1, 3, 5, 7, 9], [2, 4, 6, 8, 10])))
print(list(map(lambda x, y : [x, y], [1, 3, 5, 7, 9], [2, 4, 6, 8, 10])))


def make_repeat(n):
    return lambda s: s * n


double = make_repeat(2)
print(double(8))
print(double('FishC'))

